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ra1l [238]
3 years ago
9

Could ethanol vapor collect in low spots?

Chemistry
1 answer:
Leviafan [203]3 years ago
6 0

Answer:

Yes

Explanation:

Denatured ethanol fuel is a polar solvent, which is soluble in water. A

Polar solvent is a compound with a charge separation in chemical bonds, such as  alcohol, most acids, or ammonia. These have affinity with water and  will dissolve easily. Denatured fuel ethanol has a flash point of  -5 ° F and a vapor density of 1.5, indicating that it is heavier than air.

Consequently, ethanol vapors do not rise, similar to the gasoline vapors they are looking for  lower altitudes. The specific gravity of denatured fuel ethanol is 0.79, which  indicates that it is lighter than water and has a self-ignition temperature of 709 ° F and a  boiling point of 165-175 ° F. Like gasoline, the most denatured fuel,  the greatest danger of ethanol as an engine fuel component is its flammability.

It has a wider flammable range than gasoline (LEL is 3% and UEL is 19%).

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If the internal energy of a system increases but there is no change in temperature, then the system's energy is increasing.
777dan777 [17]

Answer:

kintic

Explanation:

4 0
3 years ago
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Which of the following is not the same as 2.97 milligrams?
Len [333]
Hi There!!!!!!

Which of the following is not the same as 2.97 milligrams?A)0.297 cg
B)0.00297 g
C) 0.0000297 KG

Answer:0.00297 g

5 0
2 years ago
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When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?
Galina-37 [17]

That will make a gold-202 nucleus.

<h3>Explanation</h3>

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the \displaystyle ^{202}_{\phantom{2}80}\text{Hg} with a neutron ^{1}_{0}n. The neutron will add 1 to the mass number 202 of ^{202}_{\phantom{2}80}\text{Hg}. However, the atomic number will stay the same.

  • New mass number: 202 + 1 = 203.
  • Atomic number is still 80.

^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}.

Double check the equation:

  • Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
  • Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The ^{203}_{\phantom{2}80}\text{Hg} nucleus loses a proton ^{1}_{1}p. Both the mass number 203 and the atomic number will decrease by 1.

  • New mass number: 203 - 1 = 202.
  • New atomic number: 80 - 1 = 79.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p.

Double check the equation:

  • Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
  • Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.

A gold-202 nucleus is formed.

6 0
3 years ago
how many grams of calcium chloride can be prepared from 60.4 G of calcium oxide and 69.0 G of hydrochloric acid and a double dis
Ksenya-84 [330]
Balanced equation: 
<span>CaO + 2 HCl --> CaCl2 + H2O </span>
<span>Calculate moles of each reactant: </span>
<span>60.4 g CaO / 56.08 g/mol = 1.08 mol CaO </span>
<span>69.0 g HCl / 36.46 g/mol = 1.89 mol HCl </span>

<span>Identify the limiting reactant: </span>
<span>Moles CaO needed to react with all HCl: </span>
<span>1.89 mol HCl X (1 mol CaO / 2 mol HCl) = 0.946 mol CaO </span>
<span>Because you have more CaO than that available, HCl is the limiting reactant. </span>

<span>Calculate moles and mass CaCl2: </span>
<span>1.89 mol HCl X (1 mol CaCl2 / 2mol HCl) X 111.0 g/mol = 105 g CaCl2</span>
8 0
3 years ago
Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of 0.0067·Ms−1: 2N2
choli [55]

Answer:

5.6 seconds

Explanation:

The reaction follows a zero-order in dinitrogen monoxide

Rate = k[N20]^0 = change in concentration/time

[N20]^0 = 1

Time = change in concentration of N2O/k

Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol

Initial concentration = moles/volume = 0.3/4 = 0.075

Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol

Concentration after t seconds = 0.15/4 = 0.0375 M

Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M

k = 0.0067 M/s

Time = 0.0375/0.0067 = 5.6 s

4 0
3 years ago
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